Computing the Link Center of a Simple Polygon
Computing the Link Center of a Simple Polygon
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To quickly assess the difficulty of the text, read a short excerpt:
Any two such neighbor- hoods, if they have non-empty intersection, intersect in a cell (because they are link con- vex). Thus by Lemma 8 it suffices to show that the assumptions of this lemma imply that iVi + i(x)n^* + iO')n^i +, (2)*0forx, y, 2€S.
Let x, y^ be three points in S and consider the neighborhoods ^^ + i(x), iV^ + i(y), and N^iz). U Ni, ^i(x)r\Ni, ^i{y)nN^{z)^0, then obviously also Ni, ^iix)nN^ + i(y)nN, ^i(z)*0, as desired.
So assume that Ni, ^. I(. X)r\N)i + i{y)nNi, {z) = 0. The ...assumptions of this lemma imply that ^* + i(^)'^-^* + iCy)*25. Because of link convexity (Lemma 2) this intersetion must be sim- ply connected. Thus Ni, + i{x}UNii + i(y) forms a subpolygon of P that is simple, i. E. It has no holes. By the assumptions of this lemma Ni, {z) intersects both 7V^ + i(x) and Ni, + i{y). Since we assume that Ni, + i{x)nNi, + i(y)nNi, {z) = 0, it follows that iV^(z) intersects Nit + i(x)UNi, + i(y) in at least two different connected components. Thus the complement of N|, + l{x)UN|, ^.
What to read after Computing the Link Center of a Simple Polygon?
You can find similar books in the "Read Also" column, or choose other free books by W Lenhart to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
Any two such neighbor- hoods, if they have non-empty intersection, intersect in a cell (because they are link con- vex). Thus by Lemma 8 it suffices to show that the assumptions of this lemma imply that iVi + i(x)n^* + iO')n^i +, (2)*0forx, y, 2€S.
Let x, y^ be three points in S and consider the neighborhoods ^^ + i(x), iV^ + i(y), and N^iz). U Ni, ^i(x)r\Ni, ^i{y)nN^{z)^0, then obviously also Ni, ^iix)nN^ + i(y)nN, ^i(z)*0, as desired.
So assume that Ni, ^. I(. X)r\N)i + i{y)nNi, {z) = 0. The ...assumptions of this lemma imply that ^* + i(^)'^-^* + iCy)*25. Because of link convexity (Lemma 2) this intersetion must be sim- ply connected. Thus Ni, + i{x}UNii + i(y) forms a subpolygon of P that is simple, i. E. It has no holes. By the assumptions of this lemma Ni, {z) intersects both 7V^ + i(x) and Ni, + i{y). Since we assume that Ni, + i{x)nNi, + i(y)nNi, {z) = 0, it follows that iV^(z) intersects Nit + i(x)UNi, + i(y) in at least two different connected components. Thus the complement of N|, + l{x)UN|, ^.
What to read after Computing the Link Center of a Simple Polygon?
You can find similar books in the "Read Also" column, or choose other free books by W Lenhart to read onlineMoreLess
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